Question

If $\alpha ,\beta ,\gamma $ are three real numbers and $A = \left[ {\begin{array}{*{20}{c}}
  1&{\cos \left( {\alpha - \beta } \right)}&{\cos \left( {\alpha - \gamma } \right)} \\
  {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\beta - \gamma } \right)} \\
  {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1
\end{array}} \right]$, then which of the following is not true.
a. $A$ is singular
b. $A$ is symmetric
c. $A$ is orthogonal
d. $A$ is not invertible

Answer 1

Hint:
We will rearrange the elements of the matrix by using the property $\cos \left( { - x} \right) = \cos x$. Then, compare the elements of the matrix with that of a symmetric matrix. Here, we will see if ${a_{12}} = {a_{21}}$, ${a_{23}} = {a_{32}}$ and ${a_{13}} = {a_{31}}$. If all such elements are equal, then the matrix is a symmetric matrix.

Complete step by step solution:
We are given a matrix $A = \left[ {\begin{array}{*{20}{c}}
  1&{\cos \left( {\alpha - \beta } \right)}&{\cos \left( {\alpha - \gamma } \right)} \\
  {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\beta - \gamma } \right)} \\
  {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1
\end{array}} \right]$
Here, we know that $\cos \left( { - x} \right) = \cos x$
We can write the elements of the matrix as
\[A = \left[ {\begin{array}{*{20}{c}}
  1&{\cos \left( { - \left( {\beta - \alpha } \right)} \right)}&{\cos \left( { - \left( {\gamma - \alpha } \right)} \right)} \\
  {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( { - \left( {\gamma - \beta } \right)} \right)} \\
  {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1
\end{array}} \right]\]
Which is equal to
\[A = \left[ {\begin{array}{*{20}{c}}
  1&{\cos \left( {\beta - \alpha } \right)}&{\cos \left( {\gamma - \alpha } \right)} \\
  {\cos \left( {\beta - \alpha } \right)}&1&{\cos \left( {\gamma - \beta } \right)} \\
  {\cos \left( {\gamma - \alpha } \right)}&{\cos \left( {\gamma - \beta } \right)}&1
\end{array}} \right]\]
Also, it is known that the matrix is symmetric when $\left[ {{a_{ij}}} \right] = \left[ {{a_{ji}}} \right]$
Then, we can say that the given matrix is a symmetric matrix.

Hence, option b is correct.

Note:
A matrix is said to be invertible if the determinant is non zero and the matrix is said to be singular if the determinant is zero. A matrix is said to be symmetric matrix when $\left[ {{a_{ij}}} \right] = \left[ {{a_{ji}}} \right]$ for every element of the matrix.